Evaluating $ \int_{0}^{\infty}\frac{\ln (1+16x^2)}{1+25x^2}\mathrm d x$

Question

how to solve such type of definite integration? I would like to see various methods to evaluate following integral $$ \int_{0}^{\infty}\frac{\ln (1+16x^2)}{1+25x^2}\mathrm d x$$

Answer 1

Consider Following parametric Integral $$I(\alpha )=\int_{0}^{\infty}\frac{\ln (1+\alpha ^2x^2)}{1+25x^2}\mathrm d x$$ We have $\displaystyle I(0)=0$ and $I(4)$ is required integral

By differentiating under integral sign (wrt $ \alpha $) we get $$I'(\alpha )=\int_{0}^{\infty}\frac{2\alpha x^2}{(1+25x^2)(1+\alpha^2x^2)}\mathrm d x=\frac{\pi}{5(\pi+\alpha )}$$

Integrating back wrt $\alpha$ we get $$I(\alpha )=\frac{1}{5} \pi \log (\alpha +5)+c$$

Since we have $\displaystyle I(0)=0\implies c=-\frac{1}{5} \pi \log (5)$

$$I(\alpha )=\frac{1}{5} \pi \log (\alpha +5)-\frac{1}{5} \pi \log (5)$$ By putting $\alpha =4$ we get required integral, $$I(4)=\frac{1}{5} \pi \log (9)-\frac{1}{5} \pi \log (5)=\frac{1}{5} \pi \log \left(\frac{9}{5}\right)$$

$$\large\int_{0}^{\infty}\frac{\ln (1+16x^2)}{1+25x^2}\mathrm d x=\frac{1}{5} \pi \log \left(\frac{9}{5}\right)$$

Answer 2

Rewrite the integral as

$$ 4 \log{2}\int_0^{\infty} dx \frac{1}{1+25 x^2} + \int_0^{\infty} dx \frac{\log {\left(\frac1{16}+x^2 \right)}}{1+25 x^2} $$

Consider

$$I(a) = \int_0^{\infty} dx \frac{\log{(a+x^2)}}{1+25 x^2} $$

$$\begin{align}I'(a) &= \int_0^{\infty} dx \frac{1}{(a+x^2)(1+25 x^2)}\\ &= \frac12 \int_{-\infty}^{\infty} dx \frac{1}{(a+x^2)(1+25 x^2)} \\ &= \frac1{4 \pi} \int_{-\infty}^{\infty}dk \, \frac{\pi}{\sqrt{a}} \, e^{-|k|\sqrt{a}} \frac{\pi}{5} e^{-|k|/5}\\ &= \frac{\pi}{10 \sqrt{a}} \int_0^{\infty} dk \, e^{-\left (\sqrt{a}+\frac15 \right ) k}\\ &= \frac{\pi}{10 \sqrt{a}} \frac1{\sqrt{a}+\frac15}\end{align}$$

Thus $$I(a)= \frac{\pi}{2} \int \frac{da}{5 a + \sqrt{a} } = \pi \int \frac{dy}{5 y+1} = \frac{\pi}{5} \log{\left (\sqrt{a}+\frac15 \right )} + C$$

We find the constant of integration by evaluating

$$I(0) = 2 \int_0^{\infty} dx \frac{\log{x}}{1+25 x^2} = -\frac{\pi}{5} \log{5} \implies C=0$$

The integral is therefore

$$\frac{2 \pi \log{2}}{5} + \frac{\pi}{5} \log{\frac{9}{20}} = \frac{\pi}{5} \log{\frac{9}{5}}$$

Answer 3

$$\frac12\int^\infty_{-\infty}\frac{\ln(1+16x^2)}{1+25x^2}{\rm d}x=2\pi i \operatorname*{Res}_{z=i/5}\frac{\ln(1-4iz)}{1+25z^2}=2\pi i\frac{\ln(1-4i^2/5)}{25(2i/5)}=\boxed{\Large\color{red}{{\dfrac{\pi}{5}\ln{\dfrac{9}{5}}}}}$$

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Note:

1. The imaginary part is odd and vanishes over a symmetric interval.
   
2. The integral over the arc vanishes as $\displaystyle \Theta\left(\frac{\ln{R}}{R}\right)$

Answer 4

Let’s consider $a$ as a parameter in a more general integral, then differentiate w.r.t.$a$ $$ I(a)=\int_0^{\infty} \frac{\ln \left(1+a x^2\right)}{1+b x^2} d x $$ and get $$ \begin{aligned} I^{\prime}(a) & =\int_0^{\infty} \frac{x^2}{\left(1+a x^2\right)\left(1+b x^2\right)} d x \\ & =\frac{1}{a-b} \int_0^{\infty}\left(\frac{1}{1+b x^2} \frac{1}{1+a x^2}\right) d x \\ & =\frac{1}{a-b}\left[\frac{1}{\sqrt{b}} \tan ^{-1}(\sqrt{b} x)-\frac{1}{\sqrt{a}} \tan ^{-1}(\sqrt{a} x)\right]_0^{\infty} \\ & =\frac{\pi}{2(a-b)}\left(\frac{1}{\sqrt{b}}-\frac{1}{\sqrt{a}}\right) \\ & =\frac{\pi}{2 \sqrt{a b}}\left(\frac{\sqrt{a}-\sqrt{b}}{a-b}\right) \\ & =\frac{\pi}{2 \sqrt{b}} \cdot \frac{1}{\sqrt{a}(\sqrt{a}+\sqrt{b})} \end{aligned} $$

Integrating back from $x=0 $ to $x=a$, $$ \begin{aligned} I(a) & =\frac{\pi}{2 \sqrt{b}} \int_0^a \frac{1}{\sqrt{x}(\sqrt{x}+\sqrt{b})} d x \\ & =\frac{\pi}{\sqrt{b}} \int_0^a \frac{d(\sqrt{x})}{\sqrt{x}+\sqrt{b}} \\ & =\frac{\pi}{\sqrt{b}}[\ln (\sqrt{x}+\sqrt{b})]_0^a\\&=\frac{\pi}{\sqrt{b}} \ln \left(1+\sqrt{\frac{a}{b}}\right) \end{aligned} $$

In particular, $$ \int_0^{\infty} \frac{\ln \left(1+16 x^2\right)}{1+25 x^2} =\frac{\pi}{\sqrt{25}} \ln \left(1+\sqrt{\frac{16}{25}}\right) =\frac{\pi}{5} \ln \left(\frac{9}{5}\right) $$

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{{\displaystyle #1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\sr}[2]{\,\,\,\stackrel{{#1}}{{#2}}\,\,\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align*} & \color{#44f}{\int_{0}^{\infty}{\ln\pars{1 + 16x^{2}} \over 1 + 25x^{2}}\dd x} \sr{x\ =\ s/4}{=} \left.{2 \over 25}\int_{-\infty}^{\infty}{\ln\pars{1 + s^{2}} \over s^{2} + \alpha^{2}}\dd s\right\vert_{\alpha\ \equiv\ 4/5} \\[5mm] = & \ {4 \over 25}\,\Re\int_{-\infty}^{\infty}{\ln\pars{1 + s\ic} \over s^{2} + \alpha^{2}}\dd s \sr{1 + s\ic\ \mapsto\ s}{=} -\,{4 \over 25}\,\Im\int_{1\ -\ \infty\ic}^{1\ +\ \infty\ic}{\ln\pars{s} \over \pars{s - 1}^{2} - \alpha^{2}}\dd s \\[5mm] & \mbox{In order to perform a closed contour integration, I'll add} \\ & \mbox{-to the integration path- a big arc of radius}\ R \to \infty\ \mbox{in the region} \\ & \braces{z\ \mid\ \Re\pars{z} > 1 } . \mbox{The arc contribution}\ vanishes\ out.\ \mbox{Therefore,} \\[5mm] & \color{#44f}{\int_{0}^{\infty}{\ln\pars{1 + 16x^{2}} \over 1 + 25x^{2}}\dd x} \\[5mm] = & \ -\,{4 \over 25}\,\Im\braces{-2\pi\ic\,\on{Res}\bracks{{\ln\pars{s} \over \pars{s - 1}^{2} - \alpha^{2}},s = 1 + \alpha}} \\[5mm] = & \ \left.{4 \over 25}\, 2\pi\,{\log\pars{1 + \alpha} \over 2\alpha}\right\vert_{\alpha\ \equiv\ 4/5} = \bbx{\color{#44f}{{\pi \over 5}\ln\pars{9 \over 5}}} \approx 0.3693 \end{align*}